Question

how many grams of copper are required to replace 4.00 grams of silver nitrate which are dissolved in water

Answer 1

Answer:

0.7482 g grams of copper are required .Explanation:

$Cu(s)+2AgNO_3(aq)\rightarrow Cu(NO_3)_2(aq)+2Ag(s)$

Moles of silver nitrate =$\frac{4.00 g}{169.87 g/mol}=0.0235474 mol$

According to reaction , 2 moles of silver nitrate reacts with 1 mol of copper.

Then 0.0235 mol of silver nitrate will react with:

$\frac{1}{2}\times 0.0235474 mol=0.0117737 mol$ of copper

Mass of 0.0117737 mol of copper = 0.117737 mol × 63.55 g/mol = 0.7482 g

0.7482 g grams of copper are required .

Answer 2

Cu(s) + 2 AgNO3 = Cu(NO3)2 + 2 Ag 63.5 g Cu ---------------- 2 x 169.87 g AgNO3 ( mass Cu ?) -------------- 4.00 g AgNO3 mass Cu = 4.00 x 63.5 / 2 x 169.87 mass Cu = 254 / 339.74 = 0.747 g of Cu


Hope this helps